Consider the polar curve $r=1+\cos(\theta)$ over the interval $\left( 0,\pi \right)$. At which value of $\theta$ does the graph of $r$ have a vertical tangent line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\pi}{3}$ (Choice B) B $\dfrac{2\pi}{3}$ (Choice C) C $\dfrac{\pi}{2}$ (Choice D) D $\dfrac{5\pi}{6}$
A vertical line has an undefined slope. The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Once we have an expression for the slope of the tangent line, we can look for the $\theta$ -values that make $\dfrac{dx}{d\theta}=0$ but don't make $\dfrac{dy}{d\theta}=0$ (because then $\dfrac{dy}{dx}$ will be indeterminate). For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={(1+\cos(\theta))}\cos(\theta) \\\\ &=\cos(\theta)+\cos^2(\theta) \\\\ \\\\ y&={(1+\cos(\theta))} \sin(\theta) \\\\ &=\sin(\theta)+\cos(\theta)\sin(\theta) \\\\ &=\sin(\theta)+\dfrac{1}{2}\sin(2\theta) \end{aligned}$ Let's find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$. $\begin{aligned} y(\theta)&=\sin(\theta)+\dfrac{1}{2}\sin(2\theta) \\\\ \dfrac{dy}{d\theta}&=\cos(\theta)+\dfrac{1}{2}\cos(2\theta)(2) \\\\ &=\cos(\theta)+\cos(2\theta) \\\\ \\\\ x(\theta)&=\cos(\theta)+\cos^2(\theta) \\\\ \dfrac{dx}{d\theta}&=-\sin(\theta)+2\cos(\theta)(-\sin(\theta) \\\\ &=-\sin(\theta)-2\cos(\theta)\sin(\theta) \end{aligned}$ Now let's solve $\dfrac{dx}{d\theta}=0$ on the interval $\left( 0,\pi \right)$. $\begin{aligned} \dfrac{dx}{d\theta}&=0 \\\\ -\sin(\theta)-2\cos(\theta)\sin(\theta)&=0 \\\\ -\sin(\theta)(1+2\cos(\theta))&=0 \\\\ \cos(\theta)=-\dfrac{1}{2}&\text{ or }\sin(\theta)=0 \end{aligned}$ Within our interval, our possible solution is $\theta=\dfrac{2\pi}{3}$. Finally, we evaluate $\dfrac{dx}{d\theta}$ for our possible value of $\theta$ and require that $\dfrac{dx}{d\theta}\ne0$. $\begin{aligned} \left.\dfrac{dy}{d\theta} \right| _{{\theta =\tfrac{2\pi }{3}}}&=\cos\left({\dfrac{2\pi}{3}}\right)+\cos\left(2\left({\dfrac{2\pi}{3}}\right)\right) \\\\ &=\cos\left({\dfrac{2\pi}{3}}\right)+\cos\left(\dfrac{4\pi}{3}\right) \\\\ &=-\dfrac{1}{2}-\dfrac{1}{2} \\\\ &=-1 \end{aligned}$ The graph of $r$ has a vertical tangent at $\theta=\dfrac{2\pi}{3}$. The graph of the tangent is shown.